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Thus we have from either method
It is evident from the last two equations that both the linear and the-angular accelerations are constant. Therefore the equations of motion are
v = •
DISCUSSION. — When / <^C ma, then v = 0, and the motion of the suspended body is about the same as that of a freely falling body. When I ^>ma then v = 0. Therefore the velocity of the falling body changes very slowly.
3. A uniform rectangular trapdoor, which is held in a vertical position, is allowed to fall. Supposing the hinges to be smooth and horizontal, find the expression for the angular velocity at any instant of the motion.
ma*' ma
2 I + ma*^
FIG. 113.
TORQUE METHOD.—The torque on the door is due to the action of its weight and the reaction of the hinges. Therefore
n           &
G = mg.~
Putting this value of G in the torque equation we get
ENERGY METHOD. — In turning through an angle 6 the door acquires a kinetic energy of f Io>2 and loses from its potential energy an amount equal to mgh. Therefore the energy equation gives
- /co2 = mgh = mg ~ (1 - cos 6).
A                                  L
Differentiating with respect to the tune
sin 8.